How do you solve #\frac { x } { 6} + 3x = \frac { 20- 2x } { 12}#?

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Answer 1 · 2016-12-11T23:25:37

#x = 1/2#

Step 1) Get each term over the same denominator - in this case #6#:

#x/6 + (6/6)*3x = (2(10 - x))/(2*6)#

#x/6 + (18x)/6 = (cancel(2)(10 - x))/(cancel(2)*6)#

#x/6 + (18x)/6 = (10 - x)/6#

#(19x)/6 = (10 - x)/6#

Step 2) Multiply each side of the equation by #6# to eliminate the fraction and keep the equation balanced:

#6*(19x)/6 = 6*(10 - x)/6#

#cancel(6) * (19x)/cancel(6) = cancel(6)*(10 - x)/cancel(6)#

#19x = 10 - x#

Step 3) Solve for #x#:

#19x + x = 10 - x + x#

#20x = 10 - 0#

#20x = 10#

#(20x)/20 = 10/20#

#(cancel(20)x)/cancel(20) = 1/2#

#x = 1/2#